Latest questions in this category.
Co³⁺ with 4 NH₃ and 2 H₂O (neutral) forms [Co(NH₃)4(H₂O)2]^{3+, balanced by 3 Cl^-, giving [Co(NH₃)4(H₂O)2]Cl₃ . This follows from NCERT principle where relation explains outcome clearly for students.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
[Co(NH₃)5(SCN)]^{2+ has the ambidentate ligand SCN^-, which can bind via S or N, showing linkage isomerism. This follows from NCERT principle where relation explains outcome clearly for students.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
[Au(CN)2]^{- is used in gold electroplating for smooth deposition, unlike the others. This follows from NCERT principle where relation explains outcome clearly for students.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: The initial rate of a reaction is 1.6 × 10⁻³mol L^{-1 s^{-1 when [A] = 0.4 M and [B] = 0.1 M. If the rate law is Rate = k[A]², what is k in L mol^{-1 s^{-1 ? These values define the system as per NCERT data. Formula: k = fracRate[A]² = frac1.6 × 10⁻³(0.4)² = frac1.6 × 10⁻³⁰.16 = 0.01 .. This is standard NCERT relation. Substitution & Calculation: Substituting values and simplifying step by step as per NCERT method. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: A gaseous reaction follows Rate = k p_A² with k = 0.1 atm^{-1 s^{-1 and p_A = 0.6 atm . What is the rate in atm s^{-1 ? These values define the system as per NCERT data. Formula: Rate = k p_A² = 0.1 × (0.6)² = 0.1 × 0.36 = 0.036 .. This is standard NCERT relation. Substitution & Calculation: Substituting values and simplifying step by step as per NCERT method. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: What is the molarity of a solution prepared by dissolving 4 g of NaOH in 500 mL of solution? (Molar mass: NaOH = 40 g/mol ) These values define the system as per NCERT data. Formula: Moles of NaOH = 4/40 = 0.1 mol. This is standard NCERT relation. Substitution & Calculation: Volume in liters = 500/1000 = 0.5 L . Molarity = 0.1/0.5 = 0.2 M . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: What is the EMF of a ll with Ni(s) | Ni²⁺(0.005 M) || Ag⁺(0.001 M) | Ag(s) at 298 K, given E_{Ni^{2+/Ni⁰ = -0.25 V and E_{Ag^{+/Ag⁰ = 0.80 V ? These values define the system as per NCERT data. Formula: E_{ll⁰ = 0.80 - (-0.25) = 1.05 V, n = 2. This is standard NCERT relation. Substitution & Calculation: E_{ll = E_{ll⁰ - 0.059/2 log frac[Ni^{2+][Ag^{+]² . Q = 0.005/(0.001)² = 5000, log Q = 3.699 . E_{ll = 1.05 - 0.059/2 × 3.699 = 1.05 - 0.109 = 0.941 V approx 0.94 V . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
CO₂ has 2 double bonds (each with 1 sigma and 1 pi), totaling 2 pi bonds. This follows from NCERT principle where relation explains outcome clearly for students.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: The solubility of PbCl₂ in water is 1.0 × 10⁻²M. What is its K_{sp ? These values define the system as per NCERT data. Formula: PbCl₂(s) Pb²+ + 2Cl-, [Pb²+] = S, [Cl-] = 2S. This is standard NCERT relation. Substitution & Calculation: K_{sp = [Pb²+][Cl-]² = S · (2S)² = 4S³ = 4 (1.0 × 10⁻²)³ = 4 × 10⁻⁶. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: Calculate the de Broglie wavelength of an electron moving at 2.05 × __10POW₇__m s^{-1 . ( h = 6.626 × 10⁻³⁴J s, m_e = 9.1 × 10⁻³¹kg ) These values define the system as per NCERT data. Formula: lambda = h/mv = frac6.626 × 10⁻³⁴⁹.1 × 10⁻³¹ × 2.05 × __10POW₇__= 3.55 × 10⁻¹¹m .. This is standard NCERT relation. Substitution & Calculation: Substituting values and simplifying step by step as per NCERT method. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: The heat capacity of a calorimeter is 10 kJ/K. If the temperature rises from 298 K to 300 K during a reaction, what is the heat absorbed by the calorimeter? These values define the system as per NCERT data. Formula: q = C Δ T. This is standard NCERT relation. Substitution & Calculation: C = 10 kJ/K, Δ T = 300 - 298 = 2 K. So, q = 10 × 2 = 20 kJ. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: What volume of 1.5 M HCl is required to react completely with 10 g of CaCO₃? (Molar mass of CaCO₃ = 100 g/mol) These values define the system as per NCERT data. Formula: Moles of CaCO₃ = 10 / 100 = 0.1 mol. This is standard NCERT relation. Substitution & Calculation: Reaction: CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O. . Moles of HCl = 0.1 × 2 = 0.2 mol. Volume = 0.2 / 1.5 = 0.1333 L = 133.3 mL ≈ 133 mL. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
